Question

suppose that the reservoir contains 1.5L of 2.0M K2PO4. For how many hours can the reservoir provide 20mM KOH at a flow rate of 1.0 ml/min if 75% consumption of K+ in the reservoir is feasible

Answer 1

Ans. Part A:

Number of moles of K2PO4 in the reservoir = Molarity x volume of solution in Liter

                                                = 2.0 M x 1.5 L

                                                = 3.0 moles

Calculate moles of K⁺ ions:        

                        K2PO4(aq) + 2 H2O(l) ---> 2 KOH(aq) + H2PO4(aq)

Stoichiometry 1: 1 mol K2PO4 forms 2 moles KOH

So, moles of KOH in the reservoir = 2 x moles of K2PO4

                                                = 2 x 3.0 moles = 6.0 moles.

Stoichiometry 2: 1 mol KOH gives 1 mol K⁺ ions.

So, moles of K⁺ in the reservoir = moles of KOH = 6.0 moles

Part B: Given, maximum K⁺ to be supplied from reservoir = 75 % of total K⁺ in the reservoir

                                                            = 75 % of 6.0 moles

                                                            = 4.5 moles

Flowrate = 20 mM KOH at flowrate 1.0 mL/ min

            Moles of KOH supplied / min =( Molarity x volume of solution in Liter ) / min

                                                         = (20 x 10^-3 M x 0.001 L) / min

                                                         = 2.0 x 10^-7 moles /min

Thus, flowrate = 2.0 x 10^-7 moles /min

Time taken to consume K+ ion reservoir = Total moles of available K+ / flowrate

                                                = 4.5 moles / (2.0 x 10^-7 moles /min)

                                                = 2.25 x 10⁷ min

                                                = 3.75 x 10⁵ hours

Note: 1 mM = 10^-3 M; 1 mL = 10^-3 L